Thread: Proving an identity

Establish the identity:

(cotθ / 1- tanθ) + (tanθ / 1 – cotθ) = 1+ tanθ + cotθ

Hello xhazel06x

Originally Posted by xhazel06x

Establish the identity:

(cotθ / 1- tanθ) + (tanθ / 1 – cotθ) = 1+ tanθ + cotθ

If we let $\displaystyle \tan\theta = t$, then

$\displaystyle \frac{\cot\theta}{1-\tan\theta}+\frac{\tan\theta}{1-\cot\theta}= \frac{\frac{1}{t}}{1-t}+\frac{t}{1-\frac{1}{t}}$

$\displaystyle =\frac{\frac{1}{t}}{1-t}+\frac{t^2}{t-1}$

$\displaystyle =\frac{\frac{1}{t}}{1-t}-\frac{t^2}{1-t}$

$\displaystyle =\frac{\frac{1}{t}-t^2}{1-t}$

$\displaystyle =\frac{\frac{1}{t}(1-t^3)}{1-t}$

$\displaystyle =\frac{\frac{1}{t}(1-t)(1+t+t^2)}{1-t}$

$\displaystyle =\frac{1}{t} + 1 + t$

$\displaystyle = 1 + \tan\theta+\cot\theta$

Grandad

Hello, xhazel06x!

Another approach . . . longer than Grandad's

Establish the identity: .$\displaystyle \frac{\cot\theta}{1-\tan\theta} + \frac{\tan\theta}{1-\cot\theta} \:=\: 1+ \tan\theta + \cot\theta$

The left side is: .$\displaystyle \frac{\dfrac{\cos\theta}{\sin\theta}}{1 - \dfrac{\sin\theta}{\cos\theta}} + \frac{\dfrac{\sin\theta}{\cos\theta}} {1 - \dfrac{\cos\theta}{\sin\theta}} $


Multiply the fractions by $\displaystyle \frac{\sin\theta\cos\theta}{\sin\theta\cos\theta}$

$\displaystyle \frac{\sin\theta\cos\theta\left(\dfrac{\cos\theta} {\sin\theta}\right)}{\sin\theta\cos\theta\left(1 - \dfrac{\sin\theta}{\cos\theta}\right)} + \frac{\sin\theta\cos\theta\left(\dfrac{\sin\theta} {\cos\theta}\right)} {\sin\theta\cos\theta\left(1 - \dfrac{\cos\theta}{\sin\theta}\right)}$ .$\displaystyle = \;\frac{\cos^2\!\theta}{\sin\theta\cos\theta - \sin^2\!\theta} + \frac{\sin^2\!\theta}{\sin\theta\cos\theta - \cos^2\!\theta} $

. . $\displaystyle =\;\frac{\cos^2\!\theta}{\sin\theta(\cos\theta-\sin\theta)} + \frac{\sin^2\!\theta}{-\cos\theta(\cos\theta - \sin\theta)}$ .$\displaystyle = \;{\color{blue}\frac{\cos\theta}{\cos\theta}}\cdot \frac{\cos^2\!\theta}{\sin\theta(\cos\theta - \sin\theta)} \;- \;{\color{blue}\frac{\sin\theta}{\sin\theta}}\cdot \frac{\sin^2\!\theta}{\cos\theta(\cos\theta - \sin\theta)} $

. . $\displaystyle = \;\frac{\cos^3\!\theta}{\sin\theta\cos\theta(\cos\ theta-\sin\theta)} - \frac{\sin^3\!\theta}{\sin\theta\cos\theta(\cos\th eta - \sin\theta)}$ .$\displaystyle = \;\frac{\cos^3\!\theta - \sin^3\theta}{\sin\theta\cos\theta(\cos\theta - \sin\theta)} $

. . $\displaystyle = \;\frac{(\cos\theta-\sin\theta)(\cos^2\!\theta + \cos\theta\sin\theta + \sin^2\!\theta)}{\sin\theta\cos\theta(\cos\theta - \sin\theta)}$ .$\displaystyle = \;\frac{\cos^2\!\theta + \cos\theta\sin\theta + \sin^2\!\theta}{\sin\theta\cos\theta} $

. . $\displaystyle =\; \frac{\cos^2\!\theta}{\sin\theta\cos\theta} + \frac{\cos\theta\sin\theta}{\sin\theta\cos\theta} + \frac{\sin^2\!\theta}{\sin\theta\cos\theta}$ .$\displaystyle =\; \frac{\cos\theta}{\sin\theta} + 1 + \frac{\sin\theta}{\cos\theta}$


. . $\displaystyle =\;\cot\theta + 1 + \tan\theta$